范围
RippleDrawableselector
在item
里面。有用。
<?xml version="1.0" encoding="utf-8"?>
<ripple xmlns:android="http://schemas.android.com/apk/res/android"
android:color="@color/pink_highlight">
<item
android:id="@android:id/mask"
android:drawable="@color/pink_highlight" />
<item
android:drawable="@drawable/bg_selectable_item" />
</ripple>
+
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item android:drawable="@color/pink_highlight_focus" android:state_focused="true" />
<item android:drawable="@color/pink_highlight_press" android:state_pressed="true" />
<item android:drawable="@color/pink_highlight_press" android:state_activated="true" />
<item android:drawable="@color/pink_highlight_press" android:state_checked="true" />
<item android:drawable="@android:color/transparent" />
</selector>
问题
无法更改选择器的默认状态颜色DrawableCompat.setTintList
RippleDrawable bg = (RippleDrawable)
ResourcesCompat.getDrawable(context.getResources(), R.drawable.bg_navigation_item, null);
StateListDrawable bgWrap = (StateListDrawable) DrawableCompat.wrap(bg.getDrawable(1));
DrawableCompat.setTintList(bgWrap, new ColorStateList(new int[][]{new int[]{}}, new int[]{Color.WHITE}));
//
someView.setBackground(bg);
它不会改变默认选择器的状态,其他一切正常。
解决方案
问题发生是因为 - 对着色应该如何工作的误解;-ColorStateList
最好满载;
<ripple xmlns:android="http://schemas.android.com/apk/res/android"
android:color="@color/pink_highlight">
<item>
<shape
android:shape="rectangle"
android:tint="@color/selectable_transparent_item" />
</item>
<item
android:id="@android:id/mask"
android:drawable="@color/pink_highlight" />
</ripple>
+
LayerDrawable bg = (LayerDrawable) ResourcesCompat.getDrawable(context.getResources(), R.drawable.bg_selectable_item, null);
Drawable bgWrap = DrawableCompat.wrap(bg.getDrawable(0));
DrawableCompat.setTintList(bgWrap, context.getResources().getColorStateList(R.color.selectable_white_item));
someView.setBackground(bg);